Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 249 255 ASYMPTOTIC MAXIMUM PRINCIPLE Boris Korenblum University at Albany, Department of Mathematics and Statistics Albany, NY 12222, U.S.A. Dedicated to W. K. Hayman on his 75th birthday Abstract. By definition the asymptotic maximum principle (AMP) is valid for a class K of analytic functions in the unit disk D if, whenever f K sup 0 θ<2π lim sup f(re iθ ) = sup f(z). z D Let K t (t 0) be the class of all f for which lim sup{(1 z ) 2 log f(z) } t. z 1 Then AMP is valid for K 0 and not valid for K t with t > 0. 1. Historical background and the main result It is well known (Lusin Privalov theorem) [3] that there are non-zero analytic functions in D that have zero radial limits on a subset E of D having full Lebesgue measure. (However, this cannot happen if E > 0 and at the same time E is of second Baire category.) More specifically, no matter how slowly k(r) tends to + as r 1, there always is a non-zero analytic f with log f(z) k( z ) and lim f(re iθ ) = 0 almost everywhere on [0, 2π) (see [2] for this and other similar results). Recently [1] a connection was found between such radial null sets E and the notion of k -entropy of E. The radial null sets represent only one kind of asymptotic behavior, namely convergence to 0. If the radial null sets are replaced by sets of asymptotic boundedness, then we are led to the following two general problems whose purpose is to expand the scope of the classical maximum principle. Problem 1. Given a class K, characterize those subsets E of D such that f K together with lim sup f(rζ) < imply f H. sup ζ E 2000 Mathematics Subject Classification: Primary 30H05, 30E25.
250 Boris Korenblum Problem 2 (Dual to the above). Given a subset E of D, identify the class K, in terms of maximal allowed radial growth, such that the above inequality together with f K imply f H. Definition. We say that the asymptotic maximum principle (AMP) is valid for the class K of analytic functions in the unit disk D if, whenever f K, sup 0 θ<2π lim sup f(re iθ ) = sup f(z). z D We give a solution to Problem 2 for the case E = D, i.e. for AMP. More specifically, we identify maximal radial growth of functions that implies validity of the AMP. Let K t, t 0, be the class of all f for which { lim sup (1 z ) 2 log f(z) } t. z 1 (Note that for t = 0 and a non-zero f the above inequality becomes an equality.) We are now in a position to state our main result. Theorem. AMP is valid for K 0 and not valid for K t with t > 0. Remark. A similar result holds for general subharmonic functions instead of log f(z). We emphasize here the analytic case for the sake of comparison with Lusin Privalov s theorem and other results. 2. Use of a Baire-category argument We proceed now to proving the above theorem. If t > 0 then the AMP fails for K t as the example { ( ) 2 } 1 + z f(z) = exp ci 1 z shows, where c is a suitable positive constant depending on t. assume t = 0, which means for f K 0 that and proceed to prove that { lim sup (1 z ) 2 log f(z) } = 0, z 1 C = sup ζ D lim sup f(rζ) = sup f(z). z D We therefore If the first supremum above is infinite then there is nothing to prove. We therefore suppose that C <.
Asymptotic maximum principle 251 If now A > C, we define the following sets: E r = E r,f = { ζ D : f(ϱζ) A, r ϱ < 1 }, 1 2 r < 1. It is easily seen that {E r : 1 2 r < 1} is a nested family of closed sets whose union equals D. Let G r denote the interior of E r and write G = r 1/2 G r ; ζ G means that there is some δ = δ(ζ) such that z ζ < δ, z D imply f(z) A. We now claim that the open set G is dense in D, and therefore the closed set F = D\G is nowhere dense. In fact, picking an arbitrary monotone sequence 1 2 r 1 < r 2 < 1 we see that every ζ in G ultimately is in G rn and thus not in E rn \ G rn ; on the other hand, every ζ in D is either in some G rn or in some E rn \ G rn. Therefore G D \ n (E r n \ G rn ). By the Baire category theorem, the set n (E r n \ G rn ) being a countable union of nowhere dense subsets of D must have a dense complement, which proves our claim. By the Heine Borel theorem, in order to prove our theorem it is enough to show that F =. We proceed now to show that the assumption F leads to a contradiction. Recall that F consists of those points ζ D such that for every δ > 0 there is some z in D with z ζ < δ and f(z) > A. Pick now some Ã, C < Ã < A, and construct the sets Ẽr 1, 2 r < 1, that relate to F essentially as E r relate to D: Ẽ r : { ζ F : f(ϱζ) Ã, r ϱ < 1}. As before, { Ẽ r : 1 2 r < 1} is a nested family of closed sets whose union equals F. Since F, as well as all its relatively open subsets, are of second Baire category in themselves, we can use again the Baire category argument to find that there is an open arc I D and some r 0, 1 2 r 0 < 1, such that I F Ẽr 0. Let K = (Ī \ I) (I F ). This set is a closed and nowhere dense subset of Ī such that f(ϱζ) Ã for all ζ K and r 1 ϱ < 1, where r 1 is some number between r 0 and 1. It then follows that there is a constant B > Ã such that f(ϱζ) B for all ζ K, 0 ϱ < 1. The critical step now is to extend this estimate to the whole sector S = {z D : z/ z I}. Let Ī \ K = n J n where J n are disjoint open arcs with end
252 Boris Korenblum points e iα n and e iβ n. Consider the open sector S n abutting on J n. We need to prove that for all n, f(z) B, z S n. Recall that J n G, which implies that for every closed subarc γ J n there is some number a = a(γ) < 1 so that f(ϱζ) A if ζ γ and a ϱ < 1. Construct an increasing sequence of closed arcs J n1 J n2 J n3 having a common midpoint with J n and such that J k nk = J n. For the corresponding numerical sequence a nk = a( J nk ), k = 1, 2,..., we assume that a n1 < a n2 < 1. This construction yields a piecewise constant function ψ n (θ) equal to a nk if e iθ J nk \ J n,k 1. By the construction we have f(ϱe iθ ) A if ψ n (θ) ϱ < 1, α n < θ < β n. We would like to replace ψ by a continuous function ϕ n (θ) ψ n (θ) having similar properties; this can be achieved by cutting the corners of the graph of ψ n, i.e., linearly interpolating between discontinuity points of ψ n. We thus obtain a continuous curve ϱ = ϕ n (θ), α n θ β n, joining the end points e iα n and e iβ n of Jn and satisfying f(ϱe iθ ) A if ϕ n (θ) ϱ < 1, α n < θ < β n. The required estimate f(z) B, z S n, is then a consequence of the following lemma that is in fact a sharper form of a classical Phragmén Lindelöf theorem (see [4], v.2, p. 214). 3. A Phragmén Lindelöf type lemma Lemma. Let S be an open sector bounded by an arc J = {z : z = 1, α Arg z α} and two radii R ± = {z : 0 z < 1, Arg z = ±α}. Let Γ be the subdomain of S bounded by J and a curve γ whose polar equation is ϱ = ϕ(θ) with ϕ continuous on [ α, α], 0 < ϕ(θ) < 1, α < θ < α, ϕ(±α) = 1. If f is analytic in S, continuous on R + R S, and satisfies ( ) then lim sup {(1 z ) 2 log f(z) } = 0, z 1, z S sup { f(z) : z R + R Γ } = sup { f(z) : z S }. Proof. It is convenient to change the setting from the sector S to that of the half-strip Ω = {x + iy : 1 < x < 1, y > 0}, which is done by an elementary conformal mapping. We have Ω = [ 1, 1] l l + { },
Asymptotic maximum principle 253 where l ± = {±1 + iy : y > 0}. We are going to prove a somewhat stronger version of the lemma (cf. Remark at the end of Section 1), in which log f(z) is replaced by an arbitrary subharmonic v: Ω [, ) which extends to be continuous on l l + Ω { }. (Note that continuity here is equivalent to the usual continuity of e v.) Our assumptions are: (i) there is a region Γ between [ 1, 1] and a continuous curve γ : y = ϕ(x), with ϕ( 1) = ϕ(1) = 0, ϕ(x) > 0, 1 < x < 1, such that (ii) sup { v(z) : z l l + Γ } = M < ; ( ) lim sup {y 2 v(x + iy)} = 0. y 0+, 1<x<1 Our aim is to show that sup{v(z) : z Ω} = M. First we modify v by subtracting a harmonic function that is positive in Ω. Let Ψ(x, y) = 1/2 Im z 2 = xy (x 2 + y 2 ) 2. Clearly, Ψ is harmonic and positive for x > 0, y > 0, Ψ = 0 if x = 0 or y = 0, and Ψ(x, mx) = m/(1 + m 2 ) 2 x 2. Consider now v a (z) = v(z) a ( Ψ(1 + x, y) + Ψ(1 x, y) ), a > 0. Let Ω 1 = {z = x + iy Ω : y > min(1 x, 1 + x)}. By ( ) v a (z) is bounded above on Ω 1. By applying the classical Phragmén Lindelöf theorem referred to earlier, we obtain that v a (z) is bounded above in Ω 1 ; it is also continuous on Ω 1 except, perhaps, at ±1. To make it continuous on Ω 1 we adjust it as follows with ε 1 > 0: v a,ε1 (z) = v a (z) + log (z + 1)/(z + 1 + ε 1 ) + log (1 z)/(1 + ε 1 z). We then have v a,ε1 (z) < v a (z) < v(z), z Ω 1 ; also, v a,ε1 is continuous on Ω 1, with v a,ε1 (±1) =. The next step requires the curve γ : y = ϕ(x), 1 x 1, to satisfy the following additional condition: ϕ is differentiable, with ϕ (x) 1 2 ; this does not result in any loss of generality, as ϕ(x) can be replaced by any smaller positive function. With this assumption made, we see that Γ Ω 1 = ; also, every straight
254 Boris Korenblum line with slope m, m > 1, intersects the boundary of the region Ω 2 lying between Ω 1 and Γ at most at two points. Consider now: v a,ε1,ε 2 (z) = v a,ε1 (z) a ( Ψ(x + 1 ε 2, y) + Ψ(1 ε 2 x, y) ), 0 < ε 2 < 1. This function is subharmonic in the region Ω (ε2) Ω lying between the two lines l ε ± 2 = {z = x + iy : x = ±(1 ε 2 )}; also, v a,ε1,ε 2 (z) < v a,ε1 (z) < v a (z) < v(z), z Ω (ε 2). We apply now the classical maximum principle to v a,ε1,ε 2 (z) in the region Ω (ε 2) 3 Ω (ε2) \ Γ whose boundary is made up of the following parts: B 1 : the segment of γ between the lines λ + ε 2 : y = 2(x 1 + ε 2 ) and λ ε 2 : y = 2(x + 1 ε 2 ); B 2 : the two segments of the above lines lying in Ω 2 ; B 3 : the two segments of Ω 1 lying between lε 2 and λ ε 2, and between λ + ε 2 and l ε + 2, respectively; B 4 : parts of l ε ± 2 lying in Ω 1. We thus obtain for z 0 Ω (ε 2) 3 that v a,ε1,ε 2 (z 0 ) is less than or equal to max ( max{v(z) : z B 1 }, max{v a,ε1,ε 2 (z) : z B 2 B 3 }, max{v a,ε1 (z) : z B 4 } ). Now, we let ε 2 0, holding ε 1 and a fixed. The second maximum in parentheses then tends to 0 due to ( ), and the third maximum tends to max{v a,ε1 (z) : z l + l } due to continuity of v a,ε1 on Ω 1 ; at the same time v a,ε1,ε 2 (z 0 ) v 2a,ε1 (z 0 ) and Ω (ε 2) 3 eventually includes all z 0 Ω 3 = Ω \ Γ. Thus for all z 0 Ω \ Γ v 2a,ε1 (z 0 ) max ( sup{v(z) : z γ}, sup{v(z) : z l + l } ). The right-hand side depends neither on a nor on ε 1. Letting a 0, ε 1 0 we get sup{v(z 0 ) : z 0 Ω \ Γ} max ( sup{v(z) : z γ}, sup{v(z) : z l + l } ), which obviously implies sup{v(z) : z Ω} = sup{v(z) : z l l + Γ}.
Asymptotic maximum principle 255 4. Completion of the proof of the theorem Applying the above lemma to each S n we obtain sup{ f(z) : z S n } B. Since the closure of n S n is the whole sector S we deduce that f(z) is bounded by B in S. We also have lim sup ϱ 1 f(ϱζ) C < Ã on I. On the two boundary radii R 1 and R 2 f(ϱζ) Ã < A, r 1 ϱ < 1, ζ Ī \ I. This implies that log f(z) is dominated in S by a harmonic function U(z) whose boundary values on R 1 and R 2 are log B, 0 z < r 1, log Ã, r 1 z < 1, and on I, U(ζ) = log C. To make U(z) continuous we increase its piecewise constant boundary values by linearly interpolating them between the discontinuity points. The resulting harmonic function U 1 (z) U(z) is continuous in the closed sector S, and on I U 1 (ζ) = log Ã. This implies that there is some r 2, r 1 < r 2 < 1, such that f(z) A, z S, r 2 z < 1, which means that I F =, contrary to the assumption. proves our theorem. This contradiction Acknowledgements. 1. The author thanks Charles Horowitz, Dmitry Khavinson, and Pascal Thomas for helpful criticism of the first draft of this paper. The author also thanks the referee for a careful reading of the manuscript and numerous helpful suggestions. 2. Part of this paper was done during the author s visit to the Gelbart Research Insitute for the Mathematical Sciences (Bar-Ilan University, Israel). The author thanks this institution and its director, Professor B. Pinchuk, for generous hospitality and support. References [1] Bénéteau, C., and B. Korenblum: Jensen type inequalities and radial null sets. - Analysis 21, 2001, 99 105. [2] Kahane, J. P., and Y. Katznelson: Sur le comportement radial des fonctions analytiques. - C. R. Acad. Sci. Paris Sér. A 272, 1971, 718 719. [3] Lusin, N. N., and I. T. Privalov: Sur l unicité et la multiplicité des fonctions analytiques. - Ann. Sci. École Norm. Sup. (3) 42, 1925, 143 191. [4] Markushevich, A. I.: Theory of Functions of a Complex Variable. - Chelsea Publishing Co., New York, 1977. Received 21 May 2001